A cart with mass 380 g moving on a frictionless linear air track at an initial s

Question

A cart with mass 380 g moving on a frictionless linear air track at an initial speed of 1.6 m/s undergoes an elastic collision with an initially stationary cart of unknown mass. After the collision, the first cart continues in its original direction at 0.61 m/s. What is the mass of the second cart? What is its speed after impact? What is the speed of the two-cart center of mass? Two titanium spheres approach each other head-on with the same speed and collide elastically. After the collision, one of the spheres, whose mass is 250 g, remains at rest. What is the mass of the other sphere? Assume that the initial speed of each sphere was 1.6 m/s. What is the speed of the two-sphere center of mass? m/s

Explanation / Answer

Let mass of second cart = m

       speed of second cart after impact = v

Acoording to law of conservation of momentum :-

0.38 * 1.6 + 0 = 0.38 *0.6 + mv

As, it is an elastic collision so, kinetic energy is also conserved

So, 1/2 * 0.38 *1.6*1.6 + 0 = (1/2 * 0.38 *0.6*0.6) + 1/2 * m * v2   

        0.4864 - 0.0706 = 1/2 *m *v2

           0.4158 = 1/2 *m *v2

Also, m*v = 0.38(1.6-0.61)

          m*v = 0.3762

So,v =2.21 m/sec

and , m = 170.22 gm

So, Mass of second cart = 170.22 gm

      Speed after impact = 2.21 m/sec

For, centre of mass applying conservation of momentum :-

0.38*1.6 = (0.38 + 0.17022) Vc

Vc = 1.105 m/sec

Speed of two cart centre of mass = 1.105 m/sec

For two spheres, applying conservation of momentum :-

0.25V - MV = MVx

V(0.25-M) = MVx                                                                                -1

Applying conservation of kinetic energy:-

1/2*(.25)*V*V + 1/2*M*V*V = 1/2*M*Vx*Vx                                             -2

Solving eq 1 & 2

M = 83.333 gm

Mass of other sphere = 83.33 gm

For two sphere centre of mass:-

0.25 *1.6 - 0.0833*1.6 = (0.25+0.0833)Vcen

0.4 - 0.133 = 0.333* Vcen

=0.802 m/sec

Two sphere centre of mass = 0.802 m/sec

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